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013213215X_ism02.pdf
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013213215X ism02.pdfCircuit Elements a Assessment...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
013213215X ism02.pdfCircuit Elemen...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
Page 14
2–14
CHAPTER 2. Circuit Elements
across the left branch is 5 V, so 20 +
v
9A
= 5 V, and
v
9A
=

15 V, where
v
9A
is positive at the top. Note that the current through the 20 V source must be
9 A, ﬂowing from top to bottom, and the current through the
v
g
is 6 A ﬂowing
from top to bottom. Let’s ﬁnd the power associated with the left and middle
branches:
p
9A
= (9)(

15) =

135 W
p
20V
= (9)(20) = 180 W
p
v
g
=

(6)(

4
.
5) = 27 W
p
6A
= (6)(0
.
5) = 3 W
Since there is only one component left, we can ﬁnd the total power:
p
total
=

135 + 180 + 27 + 3 +
p
ds
= 75 +
p
ds
=0
so
p
ds
must equal

75 W.
Therefore,
P
dev
=
P
abs
= 210 W
P 2.9
[a]
Yes, each of the voltage sources can carry the current required by the
interconnection, and each of the current sources can carry the voltage
drop required by the interconnection. (Note that
i
Δ
=

8 A.)
[b]
No, because the voltage drop between the top terminal and the bottom
terminal cannot be determined. For example, deﬁne
v
1
,
v
2
, and
v
3
as
shown:
The voltage drop across the left branch, the center branch, and the right
branch must be the same, since these branches are connected at the same
two terminals. This requires that
20 +
v
1
=
v
2
+ 100 =
v
3
But this equation has three unknown voltages, so the individual voltages
cannot be determined, and thus the power of the sources cannot be
determined.
P 2.10
[a]
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Page 15
Problems
2–15
[b]
V
bb
=
noload voltage of battery
R
bb
=
internal resistance of battery
R
x
=
resistance of wire between battery and switch
R
y
=
resistance of wire between switch and lamp A
R
a
=
resistance of lamp A
R
b
=
resistance of lamp B
R
w
=
resistance of wire between lamp A and lamp B
R
g
1
=
resistance of frame between battery and lamp A
R
g
2
=
resistance of frame between lamp A and lamp B
S
=
switch
P 2.11
Since we know the device is a resistor, we can use Ohm’s law to calculate the
resistance. From Fig. P2.11(a),
v
=
Ri
so
R
=
v
i
Using the values in the table of Fig. P2.11(b),
R
=

108

0
.
004
=

54

0
.
002
=
54
0
.
002
=
108
0
.
004
=
162
0
.
006
= 27 kΩ
Note that this value is found in Appendix H.
P 2.12
The resistor value is the ratio of the power to the square of the current:
R
=
p
i
2
. Using the values for power and current in Fig. P2.12(b),
5
.
5
×
10

3
(50
×
10

6
)
2
=
22
×
10

3
(100
×
10

6
)
2
=
49
.
5
×
10

3
(150
×
10

6
)
2
=
88
×
10

3
(200
×
10

6
)
2
=
137
.
5
×
10

3
(250
×
10

6
)
2
=
198
×
10

3
(300
×
10

6
)
2
=2
.
2 MΩ
Note that this is a value from Appendix H.
P 2.13
Since we know the device is a resistor, we can use the power equation. From
Fig. P2.13(a),
p
=
vi
=
v
2
R
so
R
=
v
2
p
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
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